Day 11: Gauss' Law

At the beginning of the class, we chose three circles to find the charge enclosed by the arbitrary surface and  the lines of flux in and out of the surface

In this picture, q is proportional to flux, and we figured out that the unit of constant k is c^2/N*m^2.

We also found a relationship between q/e_{o and E*A.}

The black net like device is called Faraday cage. We need to predict what would happen to the foils when the cylinder is charged.

What our predicted was right, which only the outer foils will move away from the cylinder. 

In this picture, it shows that the Q is on the surface when the cylinder is charged.

This picture shows a relationship betweeen E and  σ.

In this picture, it shows what changes in perimeter, area, and volume when the radius is double.

This picture shows the Gauss' law

In this picture, we plugged in 4πr^2 as area and found the relationship between E, q, and r.

This picture shows that we used the Gauss' law to find spherically symmetric charge distributions and the electric field.

In this activity, the professor put a ball inside a micro wave and it fires, when professor put a CD inside, it broke, a match with fire inside and nothing happened. And finally a bubble inside, the bubble fires

This picture shows a problem on Geometry of Cylinders. We found the volume in terms of p, r, and L, fraction of the charge with a radius of r/2 and total charge Q, and the surface area of the cylinder.

In these two pictures we also found the electric field about the circle.

Conclusion

In today's class, we analyzed Gauss' Law and utilize it for both electrical and gravitational fields and how they work in real life. We also analyzed the relationship between electric field give the charges of Gauss' Law with different conducting materials. We also looked at what happened when different things were put in microwave and we can tell and preferred paths of electrical discharges.

Day 10: Dipole Moments and Electric Flux

At the beginning of the class, we drew the sketch about two lines filled with positive particles and negative particles. The red arrows represent the direction of the electric field. And the electron is moving toward the positive side.

This photo shows the electric dipole in an electric field, which the direction of the electric dipole is from negative to positive. And the direction of forces of two particles are opposite.

We found that dipole moment is p = 2aq and τ = F*r Therefore, the net torque is P*E.

We also found the change in potential energy to be the work. And we got U = PE(cosθ 1- cosθ 2).

The red arrows are the direction of normal vectors of the electric dipole.

The true normal vectors of the electric dipole using vpython.

This shows the electric field lines between a positive and a negative particle and the direction of electric field is always from positive to negative.

This is normal vectors we drew on the lab manual.

Then, we began to talk about flux, and we found the unit of flux to be Nm^2/C

In this photo we find the Flux of 4 different surfaces in the cube. Since the electric field lines is from left to right, the flux of surface 3 and 4 is 0. And the flux of surface 1 and 2 can be calculated by the equation Φ = E*A.

Conclusion

In today's class, we learned about dipole moments and how to find torque of dipole. 

We used vpython to draw the direction of normal vectors of the electric dipole. Also, we learned
about electric flux of different surfaces through a cube 

Day 9: Electric Field

4 Statements

At the beginning of the class, we are told to change four statements from gravity situations to electric situation.

We combined the equation F = k*q1*q2/r^2 and E = F/q to get E = k*q/^2

This shows the electric field of a proton.

We used 4 step to find the electric field equation.

This is our prediction for the vpython code. 

This is the actual pictures when we solved for the electric field equations.

Electric Field from Two Point Charges

This is a problem of electric field from two point charges. We need to find Etot.

Here is another example problem where the point charge changed places.

E-Field Vectors from a Uniformly Charged Rod

We also did a problem of electric field vectors from a uniformly charged rod.

The first point where there is no y-axis, the Etot is about 60000 N/C

In another example problem, we have another position where we need to find Ex and Ey. We found that the Ex is 0 and Ey is 110210 N/C

Electric Field Due to a Differential Charge

We used the method of electric field due to a differential charge to calculate the Etot, which turned out to be 60000 N/C. The result is close to our first result which was also about 60000 N/C.

Conclusion

In today's class, we first predicted the electric field vectors of a proton, then we used vpython to see the actual picture. Then we practiced with few electric field problems to calculated Etot between two charges.

Day 8: Electric Charge and Force

Balloon Demo

In the beginning of the class, the professor demonstrated a demo of balloon that is rubbed with the hair and put near the glass wall and we need to predict what would happened. Our predict is that it would stick to the wall.

The second condition is that the balloon is rubbed with silk cloth, the result would be the same even though rubbing with silk cloth would cause less negative charge than rubbing with hair.

What is A Charge?

Professor Mason asked us to come out with a definition of charge to a 7 year-old kid. The definition is in the photo.

Interactions of Scotch Tape Strips

The first experiment is when we took the non-sticky side of two strips of tape and put them close to each other, and the result was that they repelled each other.

In second experiment, we taped another tape called T on top of the original tape called tape B. And when we put tape B together, they repelled as well as when we put tape T together.

This is our observation and results of the experiment.

Electric Force Law Video Analysis 

We calculated the angle sin^(-1)(X2/L) and the force

These are the graphs for F vs. d. We fit the line with the equation F-A*r^B, where B = -2, and we find our A to be 1.45*10^-5

This is the force relationship between a positive charge and a negative charge.

We did the example problem to find the force by p1 and p2.

Demonstration of Stormball

These four photos are the predictions and the actual results when turning on the stormball machine under different conditions.

Example Problem

This is another problem which we need to find the relationship between F1 and F2.

Conclusion

Today in this class, we learned what a charge is by looking at the balloon sticking to the wall, tapes repelled each other, and the storm ball experiment. We also learned that how to calculation the force between 2 things.